I don't know how to sketch this polar curve since I cannot take the square root of a negative number. I also don't know how to handle the plus/minus resulting from taking the square root.
This is a practice question for a calculus course and I should not be using complex numbers. Also, I shouldn't be using derivatives or integrals for this.
Thanks!
On
For $0 \le \theta \le \pi/2$, you correctly worry that $r$ is not real. It follows that there is no curve in the region $0 \le \theta \le \pi/2$ (i.e. the upper right quadrant).
But for, say, $\pi/2 \le \theta \le \pi$. $\sin 2\theta \le 0$ so $r$ is defined. You have $r \to 0$ at the endpoints, $r$ is a maximum at $\theta = 3\pi/4$, and so on.
You mention the plus or minus square root. You have to take the positive square root, because $r$ is always a positive number.
The condition $-\sin(2\theta)\geq0$ will give you $\sin(2\theta)\leq0$. In the domain $\theta \in [0,2\pi]\implies2\theta\in[0,4\pi]$, so the inequality has the solution $2\theta \in [\pi,2\pi] \cup [3\pi,4pi] \implies \theta \in [\pi/2,\pi]\cup[3\pi/2,2\pi]$. For $\theta \in [\pi/2,\pi], $ $r^2$ goes from 0 to 1 and back to 0 again like a petal. Similarly you will get a petal in fourth quadrant.